Apresenta-se aqui a derivação das transformadas de Hilbert listadas na Tabela 1 do verbete principal. Essas derivações ilustram técnicas diversas para efetuar a transformação.
A partir da definição
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Neste caso, parte-se da expressão (2) de definição da transformada e calcula-se diretamente:
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{\displaystyle f(x)\;=\;k\;\;\;\implies {\hat {u}}(x)\;=\;{\frac {1}{\pi }}\;\lim _{\epsilon \to 0}\left(\int _{x-{\frac {1}{\epsilon }}}^{x-\epsilon }{\frac {k}{u\;-\;x}}\;du\;+\;\int _{x+\epsilon }^{x+{\frac {1}{\epsilon }}}{\frac {k}{u\;-\;x}}\;du\right)}
u
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{\displaystyle {\hat {u}}(x)\;=\;{\frac {k}{\pi }}\cdot \lim _{\epsilon \to 0}\left[\left.\ln |u\;-\;x|\right|_{x-{\frac {1}{\epsilon }}}^{x-\epsilon }\;+\;\left.\ln |u\;-\;x|\right|_{x+\epsilon }^{x+{\frac {1}{\epsilon }}}\right]}
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k
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{\displaystyle {\hat {u}}(x)\;=\;{\frac {k}{\pi }}\cdot \lim _{\epsilon \to 0}\left[\ln |x\;-\;\epsilon \;-\;x|\;-\;\ln |x\;-\;{\frac {1}{\epsilon }}\;-\;x|\;+\;\ln |x\;+\;{\frac {1}{\epsilon }}\;-\;x|\;-\;\ln |x\;+\;\epsilon \;-\;x|\right]}
u
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k
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⋅
lim
ϵ
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0
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{\displaystyle {\hat {u}}(x)\;=\;{\frac {k}{\pi }}\cdot \lim _{\epsilon \to 0}\left[\ln |-\epsilon |\;-\;\ln |-{\frac {1}{\epsilon }}|\;+\;\ln |{\frac {1}{\epsilon }}|\;-\;\ln |\epsilon |\right]\;=\;{\frac {k}{\pi }}\cdot \lim _{\epsilon \to 0}\left[\ln |{\frac {-1}{-1}}|\right)}
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k
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[
ln
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1
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0
{\displaystyle {\hat {u}}(x)\;=\;{\frac {k}{\pi }}\cdot \lim _{\epsilon \to 0}\left[\ln(1)\right]\;=\;0}
Com troca de variáveis
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Neste caso, parte-se da expressão (4) e a tarefa é trivial:
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{\displaystyle f(x)\;=\;k\;\;\;\implies {\hat {u}}(x)\;=\;{\frac {1}{\pi }}\;\lim _{\epsilon \to 0}\int _{\epsilon }^{\infty }{\frac {k\;-\;k}{u}}\;du\;=\;0}
Exponencial complexa
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Neste caso, parte-se da expressão (4) e calcula-se diretamente:
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{\displaystyle f(x)\;=\;e^{ix}\;\;\;\implies {\hat {u}}(x)\;=\;{\frac {1}{\pi }}\;\lim _{\epsilon \to 0}\int _{\epsilon }^{\infty }{\frac {e^{i(x+u)}\;-\;e^{i(x-u)}}{u}}\;du}
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{\displaystyle {\hat {u}}(x)\;=\;{\frac {1}{\pi }}\;\lim _{\epsilon \to 0}\int _{\epsilon }^{\infty }{\frac {e^{ix}e^{iu}\;-\;e^{ix}e^{-iu}}{u}}\;du\;=\;{\frac {e^{ix}}{\pi }}\;\lim _{\epsilon \to 0}\int _{\epsilon }^{\infty }{\frac {e^{iu}\;-\;e^{-iu}}{u}}\;du}
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2
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{\displaystyle {\hat {u}}(x)\;=\;{\frac {e^{ix}}{\pi }}\;\lim _{\epsilon \to 0}\int _{\epsilon }^{\infty }{\frac {2i\cdot sin(u)}{u}}\;du\;=\;{\frac {2ie^{ix}}{\pi }}\;\lim _{\epsilon \to 0}\int _{\epsilon }^{\infty }{\frac {sin(u)}{u}}\;du}
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e
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{\displaystyle {\hat {u}}(x)\;=\;{\frac {2ie^{ix}}{\pi }}\cdot {\frac {\pi }{2}}\;=\;ie^{ix}}
Neste caso, pode-se usar a propriedade da dilatação do eixo, com a = -1:
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{\displaystyle f(x)\;=\;e^{-ix}\;\;\;\implies {\hat {u}}(x)\;=\;sgn(-1)\cdot ie^{-1\cdot ix}\;=\;-ie^{-ix}}
Funções trigonométricas
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Neste caso, parte-se das transformadas das exponenciais complexas, aplicando-se a propriedade da linearidade:
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e
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{\displaystyle f(x)\;=\;sin(x)\;\;\;\implies {\hat {u}}(x)\;=\;{\mathcal {H}}\{{\frac {e^{ix}\;-\;e^{-ix}}{2i}}\}}
f
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1
2
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H
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H
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{\displaystyle f(x)\;=\;{\frac {1}{2i}}\left({\mathcal {H}}\{e^{ix}\}\;-\;{\mathcal {H}}\{e^{-ix}\}\right)\;=\;{\frac {1}{2i}}\left(ie^{ix}\;+\;ie^{-ix}\right)\;=\;{\frac {1}{2}}\left(e^{ix}\;+\;e^{-ix}\right)\;=\;cos(x)}
Pode-se derivar da mesma forma:
f
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i
}
{\displaystyle f(x)\;=\;cos(x)\;\;\;\implies {\hat {u}}(x)\;=\;{\mathcal {H}}\{{\frac {e^{ix}\;+\;e^{-ix}}{i}}\}}
f
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=
1
2
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H
{
e
i
x
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H
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e
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2
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{\displaystyle f(x)\;=\;{\frac {1}{2}}\left({\mathcal {H}}\{e^{ix}\}\;+\;{\mathcal {H}}\{e^{-ix}\}\right)\;=\;{\frac {1}{2}}\left(ie^{ix}\;-\;ie^{-ix}\right)\;=\;{\frac {i\cdot i}{2\cdot i}}\left(e^{ix}\;-\;e^{-ix}\right)\;=\;-\;sin(x)}
H
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{\displaystyle {\mathcal {H}}\{sin(x)\}\;=\;cos(x)\;\;\;\implies {\mathcal {H}}\{cos(x)\}\;=\;-\;sin(x)}
H
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{
d
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d
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d
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;
c
o
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{\displaystyle {\mathcal {H}}\{cos(x)\}\;=\;{\mathcal {H}}\{{\frac {d}{dx}}\;sin(x)\}=\;{\frac {d}{dx}}{\mathcal {H}}\{sin(x)\}\;=\;{\frac {d}{dx}};cos(x)\;=\;-\;sin(x)}
A derivação é imediata, a partir da definição e da paridade das funções envolvidas:
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{\displaystyle f(x)\;=\;cas(x)\;=\;cos(x)\;+\;sin(x)\;\;\;\implies {\hat {u}}(x)\;=\;{\mathcal {H}}\{cos(x)\;+\;sin(x)\}\;=\;-\;sin(x)\;+\;cos(x)\;=\;sin(-x)\;+\;cos(-x)\;=\;cas(-x)}
Funções impulsivas
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Impulso unitário
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Neste caso, deve-se partir da expressão (1) de definição da transformada e calcular diretamente:
f
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x
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δ
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u
^
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−
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∞
δ
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d
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∞
∞
1
u
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x
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δ
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u
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0
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⋅
d
u
{\displaystyle f(x)\;=\;\delta (x)\;\;\;\implies {\hat {u}}(x)\;=\;{\frac {1}{\pi }}\int _{-\infty }^{\infty }{\frac {\delta (u)}{u\;-\;x}}\;du\;=\;{\frac {1}{\pi }}\int _{-\infty }^{\infty }{\frac {1}{u\;-\;x}}\;\cdot \;\delta (u\;-\;0)\cdot \;du}
∫
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∞
∞
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a
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d
x
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f
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a
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{\displaystyle \int _{-\infty }^{\infty }f(x)\cdot \delta (x\;-\;a)\cdot dx\;=\;f(a)}
u
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1
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0
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x
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1
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x
{\displaystyle {\hat {u}}(x)\;=\;{\frac {1}{\pi }}\cdot {\frac {1}{(0\;-\;x)}}\;=\;-\;{\frac {1}{\pi x}}}
Função recíproca
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Pela propriedade da transformada inversa, o resultado acima nos dá diretamente
f
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1
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x
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δ
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{\displaystyle f(x)\;=\;{\frac {1}{x}}\;\;\;\implies {\hat {u}}(x)\;=\;\pi \;\delta (x)}
f
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1
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+
a
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u
^
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x
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=
π
δ
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x
+
a
)
{\displaystyle f(x)\;=\;{\frac {1}{x\;+\;a}}\;\;\;\implies {\hat {u}}(x)\;=\;\pi \;\delta (x\;+\;a)}
Impulsos de ordem superior
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As transformadas das funções impulsivas de ordem superior podem ser calculadas a partir da propriedade da comutatividade com a diferenciação:
f
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x
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δ
n
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n
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d
x
n
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1
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−
1
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x
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|
n
>
1
{\displaystyle f(x)\;=\;\delta ^{n}(x)\;\;\;\implies {\hat {u}}(x)\;=\;{\frac {d^{n\;-\;1}}{dx^{n\;-\;1}}}\left(\;-\;{\frac {1}{\pi x}}\right)\;\;\;|\;n\;>\;1}
Funções racionais
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Denominador quadrático x2 + 1
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Nesse caso, pode-se aplicar a técnica das frações parciais:
f
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)
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1
x
2
+
1
⟹
u
^
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=
H
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1
x
2
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1
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=
H
[
1
2
(
1
1
+
i
x
+
1
1
−
i
x
)
]
{\displaystyle f(x)\;=\;{\frac {1}{x^{2}\;+\;1}}\;\;\;\implies {\hat {u}}(x)\;=\;{\mathcal {H}}\left({\frac {1}{x^{2}\;+\;1}}\right)\;=\;{\mathcal {H}}\left[{\frac {1}{2}}\left({\frac {1}{1\;+\;ix}}\;+\;{\frac {1}{1\;-\;ix}}\right)\right]}
u
^
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x
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=
1
2
[
u
^
1
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+
u
^
2
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]
|
f
1
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1
+
i
x
f
2
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=
1
1
−
i
x
{\displaystyle {\hat {u}}(x)\;=\;{\frac {1}{2}}\left[{\hat {u}}_{1}(x)\;+\;{\hat {u}}_{2}(x)\right]\;\;\;\;\;|\;f_{1}(x)\;=\;{\frac {1}{1\;+\;ix}}\;\;\;f_{2}(x)\;=\;{\frac {1}{1\;-\;ix}}}
G
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ω
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⟹
G
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1
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ω
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=
i
⋅
s
g
n
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ω
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⋅
G
1
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i
⋅
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n
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⋅
2
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e
ω
⋅
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=
−
i
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2
π
e
ω
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u
1
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−
ω
)
{\displaystyle G_{1}(\omega )\;=\;2\pi e^{\omega }\cdot u_{1}(-\omega )\;\;\;\;\;\implies {\hat {G}}_{1}(\omega )\;=\;i\cdot sgn(\omega )\cdot G_{1}(\omega )\;=\;i\cdot sgn(\omega )\cdot 2\pi e^{\omega }\cdot u_{1}(-\omega )\;=\;-\;i\cdot 2\pi e^{\omega }\cdot u_{1}(-\omega )}
u
1
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x
)
=
F
−
1
{
G
^
1
(
ω
)
}
=
F
−
1
{
−
i
⋅
2
π
e
ω
⋅
u
1
(
−
ω
)
}
=
−
i
⋅
F
−
1
{
2
π
e
ω
⋅
u
1
(
−
ω
)
}
=
−
i
⋅
1
1
+
i
x
{\displaystyle u_{1}(x)\;=\;{\mathcal {F}}^{-1}\{{\hat {G}}_{1}(\omega )\}\;=\;{\mathcal {F}}^{-1}\{-\;i\cdot 2\pi e^{\omega }\cdot u_{1}(-\omega )\}\;=\;-\;i\cdot {\mathcal {F}}^{-1}\{2\pi e^{\omega }\cdot u_{1}(-\omega )\}\;=\;-\;i\cdot {\frac {1}{1\;+\;ix}}}
G
2
(
ω
)
=
2
π
e
−
ω
⋅
u
1
(
ω
)
⟹
G
^
1
(
ω
)
=
i
⋅
s
g
n
(
ω
)
⋅
2
π
e
−
ω
⋅
u
1
(
ω
)
=
i
⋅
2
π
e
−
ω
⋅
u
1
(
ω
)
{\displaystyle G_{2}(\omega )\;=\;2\pi e^{-\omega }\cdot u_{1}(\omega )\;\;\;\;\;\implies {\hat {G}}_{1}(\omega )\;=\;i\cdot sgn(\omega )\cdot 2\pi e^{-\omega }\cdot u_{1}(\omega )\;=\;i\cdot 2\pi e^{-\omega }\cdot u_{1}(\omega )}
u
2
(
x
)
=
F
−
1
{
G
^
2
(
ω
)
}
=
i
⋅
1
1
−
i
x
{\displaystyle u_{2}(x)\;=\;{\mathcal {F}}^{-1}\{{\hat {G}}_{2}(\omega )\}\;=\;i\cdot {\frac {1}{1\;-\;ix}}}
u
(
x
)
=
1
2
[
u
1
(
x
)
+
u
2
(
x
)
]
=
1
2
[
−
i
⋅
1
1
+
i
x
+
i
⋅
1
1
−
i
x
]
=
−
x
x
2
+
1
{\displaystyle u(x)\;=\;{\frac {1}{2}}\left[u_{1}(x)\;+\;u_{2}(x)\right]\;=\;{\frac {1}{2}}\left[-\;i\cdot {\frac {1}{1\;+\;ix}}\;+\;i\cdot {\frac {1}{1\;-\;ix}}\right]\;=\;-\;{\frac {x}{x^{2}\;+\;1}}}
f
(
x
)
=
x
x
2
+
1
⟹
u
^
(
x
)
=
1
x
2
+
1
{\displaystyle f(x)\;=\;{\frac {x}{x^{2}\;+\;1}}\;\;\;\implies {\hat {u}}(x)\;=\;{\frac {1}{x^{2}\;+\;1}}}
Denominador quadrático (x2 + 1)2
editar
Usando também aqui a relação com a transformada de Fourier, seja
f
(
x
)
=
x
e
−
x
u
(
x
)
{\displaystyle f(x)\;=\;xe^{-x}u_{(}x)}
F
(
ω
)
=
1
−
ω
2
(
1
+
ω
2
)
2
−
i
2
ω
(
1
+
ω
2
)
2
{\displaystyle F_{(}\omega )\;=\;{\frac {1\;-\;\omega ^{2}}{(1\;+\;\omega ^{2})^{2}}}\;-\;i{\frac {2\omega }{(1\;+\;\omega ^{2})^{2}}}}
H
{
1
−
x
2
(
1
+
x
2
)
2
}
=
−
2
x
(
1
+
ω
2
)
2
{\displaystyle {\mathcal {H}}\left\{{\frac {1\;-\;x^{2}}{(1\;+\;x^{2})^{2}}}\right\}\;=\;\;-\;{\frac {2x}{(1\;+\;\omega ^{2})^{2}}}}
Sinais importantes em aplicações práticas
editar
Função retangular
editar
Neste caso, parte-se da expressão de definição (5):
f
(
x
)
=
r
e
c
t
(
x
)
⟹
u
^
(
x
)
=
−
1
π
∫
−
∞
∞
r
e
c
t
(
x
−
u
)
u
d
u
{\displaystyle f(x)\;=\;rect(x)\;\;\;\implies {\hat {u}}(x)\;=\;-\;{\frac {1}{\pi }}\int _{-\infty }^{\infty }{\frac {rect(x\;-\;u)}{u}}\;du}
u
^
(
x
)
=
−
1
π
∫
x
−
1
2
x
+
1
2
1
u
d
u
{\displaystyle {\hat {u}}(x)\;=\;-\;{\frac {1}{\pi }}\int _{x\;-\;{\frac {1}{2}}}^{x\;+\;{\frac {1}{2}}}\;\;\;{\frac {1}{u}}\;du}
u
^
(
x
)
|
x
>
1
2
=
−
1
π
∫
x
−
1
2
x
+
1
2
1
u
d
u
=
1
π
ln
|
u
|
|
x
−
1
2
x
+
1
2
=
1
π
ln
|
x
−
1
2
x
+
1
2
|
{\displaystyle \left.{\hat {u}}(x)\right|_{x\;>\;{\frac {1}{2}}}\;=\;-\;{\frac {1}{\pi }}\int _{x\;-\;{\frac {1}{2}}}^{x\;+\;{\frac {1}{2}}}{\frac {1}{u}}\;du\;=\;{\frac {1}{\pi }}\left.\ln |u|\right|_{x\;-\;{\frac {1}{2}}}^{x\;+\;{\frac {1}{2}}}\;=\;{\frac {1}{\pi }}\;\ln |{\frac {x\;-\;{\frac {1}{2}}}{x\;+\;{\frac {1}{2}}}}|}
u
^
(
x
)
|
0
<
x
<
1
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{\displaystyle \left.{\hat {u}}(x)\right|_{0\;<\;x\;<\;{\frac {1}{2}}}\;=\;-\;{\frac {1}{\pi }}\;\lim _{\epsilon \to 0}\left(\int _{x\;-\;{\frac {1}{2}}}^{\epsilon }\;\;\;{\frac {1}{u}}\;du\;+\;\int _{\epsilon }^{x\;+\;{\frac {1}{2}}}\;\;\;{\frac {1}{u}}\;du\right)\;=\;-\;{\frac {1}{\pi }}\;\lim _{\epsilon \to 0}\left(\ln |\epsilon |\;-\;\ln |x\;-\;{\frac {1}{2}}|\;+\;\ln |x\;+\;{\frac {1}{2}}|\;-\;\ln |\epsilon |\right)}
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{\displaystyle \left.{\hat {u}}(x)\right|_{0\;<\;x\;<\;{\frac {1}{2}}}\;=\;{\frac {1}{\pi }}\;\ln |{\frac {x\;-\;{\frac {1}{2}}}{x\;+\;{\frac {1}{2}}}}|}
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{\displaystyle \left.{\hat {u}}(x)\right|_{x\;<\;0}\;=\;sgn(-1)\;\cdot \;\left.{\hat {u}}(-1\cdot x)\right|_{x\;>\;0}\;=\;-\;{\frac {1}{\pi }}\;\ln |{\frac {-x\;-\;{\frac {1}{2}}}{-x\;+\;{\frac {1}{2}}}}|\;=\;-\;{\frac {1}{\pi }}\;\ln |{\frac {x\;+\;{\frac {1}{2}}}{x\;-\;{\frac {1}{2}}}}|\;=\;{\frac {1}{\pi }}\;\ln |{\frac {x\;-\;{\frac {1}{2}}}{x\;+\;{\frac {1}{2}}}}|}
Função seno cardinal
editar
Neste caso, calcula-se primeiro a transformada Fourier de f(x) e obtém-se o espectro de frequências de û(t):
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{\displaystyle f(x)\;=\;sinc(x)\;=\;{\frac {sin(\pi x)}{\pi x}}\;\;\;\implies {\mathcal {F}}\{{\hat {u}}(x)\}\;=\;i\cdot sgn(\omega )\cdot G(\omega )\;=\;i\cdot sgn(\omega )\cdot {\mathcal {F}}\{sinc(x)\}}
tabela de transformadas de Fourier ,
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{\displaystyle {\mathcal {F}}\{{\hat {u}}(x)\}\;=\;i\cdot sgn(\omega )\cdot \pi \cdot rect\left({\frac {\omega }{2\pi }}\right)}
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{\displaystyle {\hat {u}}(x)\;=\;{\mathcal {F}}^{-1}\left[i\cdot sgn(\omega )\cdot \pi \cdot rect\left({\frac {\omega }{2\pi }}\right)\right]\;=\;{\frac {1}{2\pi }}\int _{-\infty }^{\infty }i\cdot sgn(\omega )\cdot \pi \cdot rect\left({\frac {\omega }{2\pi }}\right)\cdot e^{i\omega x}\cdot d\omega }
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{\displaystyle {\hat {u}}(x)\;=\;{\frac {i}{2}}\left[\int _{-\infty }^{0}-\;rect\left({\frac {\omega }{2\pi }}\right)\cdot e^{i\omega x}\cdot d\omega \;+\;\int _{0}^{\infty }rect\left({\frac {\omega }{2\pi }}\right)\cdot e^{i\omega x}\cdot d\omega \right]}
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{\displaystyle {\hat {u}}(x)\;=\;{\frac {i}{2}}\left[\int _{-\pi }^{0}-\;e^{i\omega x}\cdot d\omega \;+\;\int _{0}^{\pi }e^{i\omega x}\cdot d\omega \right]\;=\;{\frac {i}{2}}\left[-\;{\frac {1}{ix}}\cdot \left.e^{i\omega x}\right|_{-\pi }^{0}\;+\;{\frac {1}{ix}}\cdot \left.e^{i\omega x}\right|_{0}^{\pi }\right]}
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{\displaystyle {\hat {u}}(x)\;=\;{\frac {1}{2x}}\left[-1\;+\;e^{-i\pi x}\;+\;e^{i\pi x}\;-\;1\right]\;=\;{\frac {cos(\pi x)\;-\;1}{x}}}
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{\displaystyle {\mathcal {F}}\{{\frac {sin(x)}{x}}\}\;=\;{\mathcal {F}}\{{\frac {sin({\frac {\pi x}{\pi }})}{\frac {\pi x}{x}}}\}\;=\;sgn\left({\frac {1}{\pi }}\right)\cdot {\frac {cos\left({\frac {\pi x}{\pi }}\right)\;-\;1}{\frac {\pi x}{\pi }}}\;=\;{\frac {cos(x)\;-\;1}{x}}}
![{\displaystyle {\hat {u}}(x)\;=\;{\frac {k}{\pi }}\cdot \lim _{\epsilon \to 0}\left[\left.\ln |u\;-\;x|\right|_{x-{\frac {1}{\epsilon }}}^{x-\epsilon }\;+\;\left.\ln |u\;-\;x|\right|_{x+\epsilon }^{x+{\frac {1}{\epsilon }}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7672a0c2780ebfc216849922fb4a23136c4938d9)
![{\displaystyle {\hat {u}}(x)\;=\;{\frac {k}{\pi }}\cdot \lim _{\epsilon \to 0}\left[\ln |x\;-\;\epsilon \;-\;x|\;-\;\ln |x\;-\;{\frac {1}{\epsilon }}\;-\;x|\;+\;\ln |x\;+\;{\frac {1}{\epsilon }}\;-\;x|\;-\;\ln |x\;+\;\epsilon \;-\;x|\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/536c045672f9cadeb8d05ff7955207cb39431959)
![{\displaystyle {\hat {u}}(x)\;=\;{\frac {k}{\pi }}\cdot \lim _{\epsilon \to 0}\left[\ln |-\epsilon |\;-\;\ln |-{\frac {1}{\epsilon }}|\;+\;\ln |{\frac {1}{\epsilon }}|\;-\;\ln |\epsilon |\right]\;=\;{\frac {k}{\pi }}\cdot \lim _{\epsilon \to 0}\left[\ln |{\frac {-1}{-1}}|\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb935f39b67d844bef29b490ed652ef850115115)
![{\displaystyle {\hat {u}}(x)\;=\;{\frac {k}{\pi }}\cdot \lim _{\epsilon \to 0}\left[\ln(1)\right]\;=\;0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68400fbea0ef7a436276ac6068a8f04bc0ed1776)
![{\displaystyle f(x)\;=\;{\frac {1}{x^{2}\;+\;1}}\;\;\;\implies {\hat {u}}(x)\;=\;{\mathcal {H}}\left({\frac {1}{x^{2}\;+\;1}}\right)\;=\;{\mathcal {H}}\left[{\frac {1}{2}}\left({\frac {1}{1\;+\;ix}}\;+\;{\frac {1}{1\;-\;ix}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/306d67ab9a28ca0512210a4f48a65a387e114090)
![{\displaystyle {\hat {u}}(x)\;=\;{\frac {1}{2}}\left[{\hat {u}}_{1}(x)\;+\;{\hat {u}}_{2}(x)\right]\;\;\;\;\;|\;f_{1}(x)\;=\;{\frac {1}{1\;+\;ix}}\;\;\;f_{2}(x)\;=\;{\frac {1}{1\;-\;ix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb05ba30ddb35cc70b4a9b099deb5aff3a732ded)
![{\displaystyle u(x)\;=\;{\frac {1}{2}}\left[u_{1}(x)\;+\;u_{2}(x)\right]\;=\;{\frac {1}{2}}\left[-\;i\cdot {\frac {1}{1\;+\;ix}}\;+\;i\cdot {\frac {1}{1\;-\;ix}}\right]\;=\;-\;{\frac {x}{x^{2}\;+\;1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1795c3d93521b8177f130c682bb80390c22fd32)
![{\displaystyle {\hat {u}}(x)\;=\;{\mathcal {F}}^{-1}\left[i\cdot sgn(\omega )\cdot \pi \cdot rect\left({\frac {\omega }{2\pi }}\right)\right]\;=\;{\frac {1}{2\pi }}\int _{-\infty }^{\infty }i\cdot sgn(\omega )\cdot \pi \cdot rect\left({\frac {\omega }{2\pi }}\right)\cdot e^{i\omega x}\cdot d\omega }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8a4304e87f9b5fa769fd796a175c3f10c62e331)
![{\displaystyle {\hat {u}}(x)\;=\;{\frac {i}{2}}\left[\int _{-\infty }^{0}-\;rect\left({\frac {\omega }{2\pi }}\right)\cdot e^{i\omega x}\cdot d\omega \;+\;\int _{0}^{\infty }rect\left({\frac {\omega }{2\pi }}\right)\cdot e^{i\omega x}\cdot d\omega \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29672b22c56bbb5cef4927a1aeaa0fb2ddbc89b2)
![{\displaystyle {\hat {u}}(x)\;=\;{\frac {i}{2}}\left[\int _{-\pi }^{0}-\;e^{i\omega x}\cdot d\omega \;+\;\int _{0}^{\pi }e^{i\omega x}\cdot d\omega \right]\;=\;{\frac {i}{2}}\left[-\;{\frac {1}{ix}}\cdot \left.e^{i\omega x}\right|_{-\pi }^{0}\;+\;{\frac {1}{ix}}\cdot \left.e^{i\omega x}\right|_{0}^{\pi }\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bfda3383001ea853b7d535338d3563d7ecc1e188)
![{\displaystyle {\hat {u}}(x)\;=\;{\frac {1}{2x}}\left[-1\;+\;e^{-i\pi x}\;+\;e^{i\pi x}\;-\;1\right]\;=\;{\frac {cos(\pi x)\;-\;1}{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6b91900d7e94c8ea826ac4c2f58b5300afbd784)
